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I don't know how to solve the integral $$ \int \exp\left(-\frac{a^2}{2x^2}-\frac{b^2x^2}{2}\right)dx$$ for $a,b \in \mathbb{R}$. My idea was to complete the square to get an expression close to the integral necessary for the error function and then a substitution which did not work. WolframAlpha also gives me a result including terms with the error function.

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  • $\begingroup$ The indefinite integral will for sure be non-standard. Maybe you can get something "readable" if you put bounds such as $\int_0^\infty$. $\endgroup$ – Zubzub Apr 7 '17 at 8:17
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It seems that it should work completing the square since $$\frac{a^2}{2x^2}+\frac{b^2x^2}{2}=\frac 12\left(bx+\frac ax\right)^2-ab$$ So let $$bx+\frac ax=u\sqrt 2\implies x=\frac{\sqrt{u^2-2 a b}+u}{\sqrt{2}\, b}\implies dx=\frac{1+\frac{u}{\sqrt{u^2-2 a b}}}{\sqrt{2}\, b}$$ So, $$\int \exp\left(-\frac{a^2}{2x^2}-\frac{b^2x^2}{2}\right)dx=\frac{e^{a b}}{\sqrt{2}\,b}\left(\int \frac{e^{-u^2} u}{\sqrt{u^2-2 a b}}\,du+\int e^{-u^2}\right)\,du$$ The second integral is trivial; for the first another change of variable $u=\sqrt{2 a b+v^2}$ leads to $$\int \frac{e^{-u^2} u}{\sqrt{u^2-2 a b}}\,du=e^{-2ab}\int e^{-v^2}\,dv=\frac{1}{2} \sqrt{\pi } e^{-2 a b} \text{erf}\left(\sqrt{u^2-2 a b}\right)$$

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